Let $A = (-4,0,6),$ $B = (-5,-1,2),$ and $C = (-6,-1,3).$  Compute $\angle ABC,$ in degrees.
From the distance formula, we compute that $AB = 3 \sqrt{2},$ $AC = \sqrt{14},$ and $BC = \sqrt{2}.$  Then from the Law of Cosines,
\[\cos \angle ABC = \frac{(3 \sqrt{2})^2 + (\sqrt{2})^2 - (\sqrt{14})^2}{2 \cdot 3 \sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}.\]Therefore, $\angle ABC = \boxed{60^\circ}.$